Answer:
![\frac{3}{7}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B7%7D)
Explanation:
Lets take the numerator of the fraction to be = x
So the denominator of the fraction is 4 more than the numerator = x+4
The fraction is ;![\frac{x}{4+x}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B4%2Bx%7D)
Now add 4 to the numerator and add 7 to the denominator as;
![\frac{x+4}{4+x+7} =\frac{x+4}{x+11}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%2B4%7D%7B4%2Bx%2B7%7D%20%3D%5Cfrac%7Bx%2B4%7D%7Bx%2B11%7D)
This new fraction is equal to 1 half =1/2
write the equation as;
![\frac{x+4}{x+11} =\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%2B4%7D%7Bx%2B11%7D%20%3D%5Cfrac%7B1%7D%7B2%7D)
perform cross-product
2(x+4 )=1( x+11 )
2x+8 = x + 11
2x-x = 11-8
x=3
The original fraction is;
![\frac{x}{4+x} =\frac{3}{3+4} =\frac{3}{7}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B4%2Bx%7D%20%3D%5Cfrac%7B3%7D%7B3%2B4%7D%20%3D%5Cfrac%7B3%7D%7B7%7D)
At entrance to the nozzle, the pressure is 0.180 MPa and the temperature is 1200 K. The kinetic energy of the gas entering the nozzle is very much smaller than ... The specific heat of the exhaust gas varies with temperature approximately as follows: ... Problem 4P: In an aircraft jet engine at takeoff, the combustion product.
Answer:
V1=5<u>ft3</u>
<u>V2=2ft3</u>
n=1.377
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=4lb*1.25ft3/lb=5<u>ft3</u>
state 2
V2=m.v2
V2=4lb*0.5ft3/lb= <u> 2ft3</u>
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/53)/ ln (2/5)
n=1.377
Answer:
q=2313.04![W/m^2](https://tex.z-dn.net/?f=W%2Fm%5E2)
T=690.86°C
Explanation:
Given that
Thickness t= 20 cm
Thermal conductivity of firebrick= 1.6 W/m.K
Thermal conductivity of structural brick= 0.7 W/m.K
Inner temperature of firebrick=980°C
Outer temperature of structural brick =30°C
We know that thermal resistance
![R=\dfrac{t}{KA}](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7Bt%7D%7BKA%7D)
These are connect in series
![R=\left(\dfrac{t}{KA}\right)_{fire}+\left(\dfrac{t}{KA}\right)_{struc}](https://tex.z-dn.net/?f=R%3D%5Cleft%28%5Cdfrac%7Bt%7D%7BKA%7D%5Cright%29_%7Bfire%7D%2B%5Cleft%28%5Cdfrac%7Bt%7D%7BKA%7D%5Cright%29_%7Bstruc%7D)
Heat transfer
![Q=\dfrac{\Delta T}{R}](https://tex.z-dn.net/?f=Q%3D%5Cdfrac%7B%5CDelta%20T%7D%7BR%7D)
So heat flux
q=2313.04![W/m^2](https://tex.z-dn.net/?f=W%2Fm%5E2)
Lets temperature between interface is T
Now by equating heat in both bricks
![\dfrac{980-T}{\dfrac{0.2}{1.6A}}=\dfrac{T-30}{\dfrac{0.2}{0.7A}}](https://tex.z-dn.net/?f=%5Cdfrac%7B980-T%7D%7B%5Cdfrac%7B0.2%7D%7B1.6A%7D%7D%3D%5Cdfrac%7BT-30%7D%7B%5Cdfrac%7B0.2%7D%7B0.7A%7D%7D)
So T=690.86°C